In the solution to a given problem, I am told that for an exponential distribution, the following is true:
Given an exponential distribution with a mean of 100 that models costs, in the context of there being a deductible of 20, the solution states that the probability that the costs are less than 150 = $F(130)=1-e^{-\frac{130}{100}} $. To be more specific, The solution suggests that there is something self evident in $P( 20\leq X \leq 150|X\geq 20)$ being equivalent to $P(X\leq 130)$. I cannot understand why this is.
This is simply the memorylessness property of the exponential distribution. We note that for $$X \sim \operatorname{Exponential}(\mu),$$ with PDF $$f_X(x) = \frac{1}{\mu} e^{-x/\mu}, \quad x > 0,$$ we have $$\Pr[d \le X \le x + d \mid X \ge d] = \frac{\int_{s=d}^{x+d} f_X(s) \, ds}{\int_{s=d}^\infty f_X(s) \, ds}.$$ The numerator is $$(1 - e^{-(x+d)/\mu}) - (1 - e^{-d/\mu}) = e^{-d/\mu}(1 - e^{-x/\mu}),$$ and the denominator is $e^{-d/\mu}$; hence the conditional probability is $$\Pr[d \le X \le x+d \mid X \ge d] = 1 - e^{-x/\mu} = \Pr[X \le x].$$